tricky math problems for 4th graders



Tricky Problems for Middle Schoolers

The Math League contests often have problems that are variations on a theme. By generalizing and understanding how to do certain categories of problems, we should be better able to handle specific instances. Besides the examples you find here, you can also try some sample contests.

		Common Classes of problems to Expect: - Rate type problems - Funny symbol problems. For example. a # b = (b+b-a)-(a-b) what is 9 # 8? - Word problems with multiple variables that require solving N equations with N unknowns. - Finding triangles or squares in some geometric shape - Averaging rates (like average mpg) - Coin problems. - Consistent set of rules problems. Some set of rules is given, find a consistent solution - Clock problems. For example, How many different AM times have digits that sum to 3? - Numbers from set of digits. For example, given 2,3,5,7,8,9 find two 3-digit number that have the biggest difference. - Logic puzzles. - Liars and Truthteller problems (sometimes called natives and visitors)

			Rate Type Problems example: 40 men can build 20 sheds in 2 hours. How long will it take for 5 men to build 10 sheds? Non-algebra aproach: Since we want to find the time given 5 men and 10 sheds, we want to convert the original expression to something similar. Start by noticing that if have half as many men, we will build only half as many sheds. 20 men can build 10 sheds in 2 hours. Now that the 10 sheds matches what we are looking for, what do I have to do with the hours if I change the men to 5. If I reduce the number of men, it will take longer to do the same work, so If I divide the number of men by 4, I need to multiply the time by 4. As a result, we get 5 men can build 10 sheds in 8 hours. Algebraic approach: The rate at which men can build sheds is a constant, so we can set up an equivalency: The rate is sheds / (men hours)* In other words sheds per man hours. Since they tell us the rate in the first part of the problem, we can construct a similar rate using x for hours. 20 sheds 10 sheds --------------------- = ---------------------- 40 men * 2 hours 5 men * X hours Solving for x we get 10 * 40 * 2 X = -------------------- = 8 hours 5 * 20 See also Dr Maths explanation for these types of problems:

	Here are some possible questions that I made up or derived from Mindtrap questions. The number of 's is the degree of difficulty. Most of these problems were asked during practice. I will add more with time. [] What is x 2x + 1 = 11 3x - 3 = 12 3x - 1 = 14 + 2x [] What is 10 million divided by 100,000? 10,000,000 / 100,000 = 100 [] Find A and B if 32 59 + AB A=0 B=8 ----- 99 [] Can you tell me 2 numbers, neither of which ends in 0, that when multiplied together equal 1000? 8 125 This becomes apparent if you do a prime factorization of 1000 and get 22255*5.